If you react 25.25 g of Na2CO3 with 50.0 mL of 6.0M HCl, answer the following questions:
Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2
question:
a) How many moles of HCl do you need to react completely with the 25.25 g of NaCO3?
b) What is limiting reagent?
You can answer a) after following the process shown in this post.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Then part b can be answered knowing that moles = M x L.
a) Well, did you hear about the chemist who was reading a book about Helium? It was so good that he just couldn't put it down! Anyway, to answer your question, we need to calculate the number of moles of Na2CO3 before we can determine how many moles of HCl are needed to react completely. The molar mass of Na2CO3 is 105.99 g/mol, so we can calculate the number of moles of Na2CO3 using the formula:
moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
Plugging in the values, we get:
moles of Na2CO3 = 25.25 g / 105.99 g/mol
Now that we know the number of moles of Na2CO3, we can use the balanced equation to determine the stoichiometry between Na2CO3 and HCl. According to the balanced equation, 2 moles of HCl react with 1 mole of Na2CO3. So, we can calculate the number of moles of HCl needed as:
moles of HCl needed = 2 * moles of Na2CO3
You can plug in the value of moles of Na2CO3 into this equation to find the answer!
b) Now, it's time to determine the limiting reagent. The limiting reagent is the one that is going to run out first and limit the amount of product that can be formed. To figure it out, you need to compare the number of moles of each reactant with the stoichiometry of the balanced equation. The reactant that produces fewer moles of product is the limiting reagent. So, calculate the number of moles of NaCl that can be formed from each reactant (Na2CO3 and HCl) using the stoichiometry in the balanced equation. The reactant that produces fewer moles of NaCl is the limiting reagent. I hope this helps, and remember, chemistry jokes might not get a reaction, but they're always a compound of fun!
To answer these questions, we need to use the given chemical equation:
Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2
a) To find the number of moles of HCl needed to react completely with 25.25 g of Na2CO3, we'll first convert the mass of Na2CO3 to moles using its molar mass.
The molar mass of Na2CO3 is:
2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol
Now, we can calculate the number of moles of Na2CO3:
moles of Na2CO3 = mass / molar mass
moles of Na2CO3 = 25.25 g / 105.99 g/mol
b) To determine the limiting reagent, we need to compare the number of moles of Na2CO3 to the number of moles of HCl.
Using the balanced equation, we can see that the stoichiometric ratio between Na2CO3 and HCl is 1:2. This means 1 mole of Na2CO3 reacts with 2 moles of HCl.
So, if we have n moles of Na2CO3, we would need (2n) moles of HCl for complete reaction.
Therefore, the number of moles of HCl needed to react completely with the Na2CO3 is:
moles of HCl = (2 * moles of Na2CO3)
Now, to find the limiting reagent, we compare the moles of Na2CO3 and HCl:
If moles of Na2CO3 < (2 * moles of Na2CO3), the limiting reagent is Na2CO3.
If moles of HCl < (2 * moles of Na2CO3), the limiting reagent is HCl.
If they are equal, neither is the limiting reagent.
Consider the coefficients of the balanced equation and calculate to find the limiting reagent.